# Non-entangled two qubit system get information about the individual particle states after CNOT

I'm trying to make a quantum computing mod for a game, (apologies if the notation is wrong I'm new to QM).

Let's say I have 2 qubits that are both $$\frac{1}{\sqrt{2}}(|0⟩ + |1⟩)$$ and I put them through a CNOT gate. The result would be

$$A \otimes B = \begin{pmatrix} \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} \end{pmatrix} \otimes \begin{pmatrix} \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} \end{pmatrix} = \begin{pmatrix} \frac{1}{2}\\ \frac{1}{2}\\ \frac{1}{2}\\ \frac{1}{2} \end{pmatrix}$$

$$\mathrm{CNOT}_{\text{matrix}} (A \otimes B) = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} \frac{1}{2}\\ \frac{1}{2}\\ \frac{1}{2}\\ \frac{1}{2} \end{pmatrix}= \begin{pmatrix} \frac{1}{2}\\ \frac{1}{2}\\ \frac{1}{2}\\ \frac{1}{2} \end{pmatrix}$$

Which I don't believe is an entangled system. Thus, would it in theory be possible to get something in terms of the |0⟩ and |1⟩ for each individual particle? For example, express the individual states of qubit 1 and 2 in the form:

$$\alpha|0⟩ + \beta|1⟩$$

(And if so, how would I determine this?)

• The quickest solution in this case is to notice that your output is the same as the input! PS you’re using direct sum notation rather than tensor product. The answers are the same in the case, but more generally, make sure you’re using the correct thing! Dec 7 '19 at 6:31

Correct, the final state is not an entangled state.

1/2(|00> + |01> + |10> + |11>)


Therefore it represents a tensor product of two qubits' state.

1/sqrt(2)(|0> + |1>) <tensorproduct_symbol> 1/sqrt(2)(|0> + |1>)


For both qubits,

alpha = beta = 1/sqrt(2)


This wouldn't have been possible if the qubits were entangled.

• So to find alpha and beta I would need to do an inverse tensor product. Could you explain how you would do that please, if you can. Dec 7 '19 at 4:20
• Apologies, I'm working on my formatting skills in order to better present my answer. But I'll try my best here. It's just like doing FOIL with the expression (ax+by)(cx+dy). Imagine x and y are your |0> and |1> states. So, we'll be left with coefficients ac, ad, bc and bd for the different terms in expanded expression. You could do the same in reverse. Another way to think about it is factoring out common coefficients from the expression. Did it help? If not, I'll try to come up with a better way to present this.
– ss09
Dec 7 '19 at 4:28
• Oh right, it's just algebra. For some reason I got tangled up in the QM notation and expected there to be some magical matrix or vector manipulation I could do. Thanks, this solved my problem :) Dec 7 '19 at 4:30
• That's great! We're all learning. Happy to help.
– ss09
Dec 7 '19 at 4:33

The fastest way I know to check if a given qubit is not entangled is to check if all its conditioned-on-others states are parallel. A conditioned-on-others state is a subset of your state vector where the other qubits are equal to some specific value.

For example, the first two entries in your column vector are the state of one qubit conditioned on the other being 0. The last two entries are the state of that same qubit conditioned on the other being 1. Notice that the first two entries (as a vector) are parallel to the second two entries (as a vector). Their dot product's magnitude is equal to the product of their magnitudes. (In this case it's easy to see they're parallel because they're literally identical.) Therefore that qubit is separable. Its state is equal to the first two entries (or the second two entries; doesn't matter) after normalization up to some irrelevant global phase factor.

Quirk's amplitude displays perform this extraction when applied to a subset of the wires: Your end state is not entangled. As you have written, the state vector after the gate is the same as the one before the gate. And your starting state is obviously not entangled.

As for the rest of your question, what would be the 'something' you are looking for in terms of 0's and 1's?